Preben Alsholm

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These are replies submitted by Preben Alsholm

@bunmipopoola Did you try all my lines?
Try copying the whole thing from my answer and paste it into a fresh worksheet. Then execute.
I just tried that very thing myself in Maple 17 (but the version shouldn't matter much).
No problem.

@bunmipopoola Did you try all my lines?
Try copying the whole thing from my answer and paste it into a fresh worksheet. Then execute.
I just tried that very thing myself in Maple 17 (but the version shouldn't matter much).
No problem.

@Gaia Now the issue with +- seems  resolved: It meant -.
You need eqt to be a polynomial not an equation. Simply remove =0.
You won't have any luck with algcurves[parametrize] though.

@traiannp If the system really is inconsistent, then that means there just is no solution!

@traiannp If the system really is inconsistent, then that means there just is no solution!

In Maple 17
solve(sys, [c,d,e,f,g,h,i,j,k,l,m,n,p]);

returns [], i.e. it didn't find any solution.

In Maple 17
solve(sys, [c,d,e,f,g,h,i,j,k,l,m,n,p]);

returns [], i.e. it didn't find any solution.

This simpler version also returns an incorrect result:

s:=HFloat(-infinity);
type(s,positive);
returns true.

Could you correct the syntax of your input
eqt := y^5+y^4 + -0.5109*y^3 + -0.0595*y^2 + 0.0000-(-3.1086e-015*u^4 + u^3 + -0.5109*u^2 + -0.0595*u) = 0;
so it parses in Maple?
It is not clear to me what you intended it to mean, so I won't try myself.

An implementation of the intended procedure could be
F:=proc(L1::listlist) local L;
    L:=ListTools:-Flatten(L1);
    1/2*abs((L[3]-L[1])*(L[6]-L[2])-(L[5]-L[1])*(L[4]-L[2]))
end proc;
    
F([[a,b], [c,d], [e,f]]);

An implementation of the intended procedure could be
F:=proc(L1::listlist) local L;
    L:=ListTools:-Flatten(L1);
    1/2*abs((L[3]-L[1])*(L[6]-L[2])-(L[5]-L[1])*(L[4]-L[2]))
end proc;
    
F([[a,b], [c,d], [e,f]]);

@Yankel Yes, it is quite empty!
Did you actually write anything or did you try to dump (somehow) an image there?

Notice that if you change the problematic derivative (which is D[2](f)(0,tau), and not D[1](f)(0,tau) as you wrote ) to D[1](f)(0,tau), then things work.

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